In relativity, if units of length contracts and time dilates then does unit of velocity or speed also change? I'm just starting to learn special relativity, and I'm having trouble with the following concept:
In relativity, units of length and time of moving frame are related to that of stationary one through $$x’=\frac{x}{\gamma}\quad \quad \text{ and }\quad \quad t’=t\times \gamma$$ respectively, where $\gamma$ is Lorentz Factor.
Does this also mean that units of velocity or speed, i.e. length/time are related as $$v’=\frac{x’}{t’}=\frac{x}{t}\times\frac{1}{\gamma^2}=\frac{v}{\gamma^2}?$$
Note: By unit, I mean scale of axes in a respective coordinate system and I am not asking about addition or subtraction of velocities, I am enquiring about mutual “scale” difference between the quantity called velocity as measured in two different frames in uniform relative motion to each other.
Why scale of length contracts and not expands while that of time dilates, i.e. expands when the two are symmetrical for Lorentz transformations! Only if length expands with dilation of time can the “scale” of velocity or speed in general and speed of light in particular can remain truly invariant, I guess.
 A: No. Velocities do transform in a non-intuitive way in special relativity, but not in the way you're describing. This is sadly a very common misunderstanding due to the fact that Relativity is usually first introduced using time-dilation and length contraction, without actually explaining under which conditions they are applicable. The best way to begin understanding the subject (and to also avoid all these "paradoxes") is to work with the Lorentz Transformations. In one dimension, if the frame $S'$ is moving with respect to $S$ with a velocity $v$, then
\begin{aligned}
x' &= \gamma\left( x - vt \right)\\
t' &= \gamma \left( t - \frac{v}{c^2}x\right)
\end{aligned}
provided that $x=x'=0$ when $t=t'=0$. Remember, though, that these are coordinates, not intervals. To find how intervals of length and intervals of time are related, we need to take differences of the coordinates, and since $v$ and $\gamma$ are constants, it's easy to show that the intervals satisfy similar equations:
\begin{aligned}
\Delta x' &= \gamma\left( \Delta x - v \Delta t \right)\\
\Delta t' &= \gamma\left( \Delta t - \frac{v}{c^2}\Delta x\right)
\end{aligned}
We can use the above equations to easily calculate how the velocities transform between the frames $S$ and $S'$. Remember that an observer in $S$ will calculate the velocity of an object to be $$u = \frac{\Delta x}{\Delta t},$$ and one in $S'$ will calculate it to be $$u' = \frac{\Delta x'}{\Delta t'}.$$
We can now divide $\Delta x'$ by $\Delta t'$ to show that:
$$u' = \frac{\Delta x'}{\Delta t'} = \frac{u - v}{1 - \frac{uv}{c^2}}.$$
Why doesn't your argument work?
Length contraction and time dilation are special cases of the general formulae that I have given above. They hold when certain conditions are satisfied, and these conditions are more certainly not satisfied simultaneously. Which means dividing the equations is not going to give you anything sensible. In special relativity, it is best to think in terms of "events" which occur at spacetime points $(t, x)$ to avoid such false "paradoxes". My answer here, and the links at the end, should explain it in more detail.
A: Along with Philip’s answer, this video ( https://youtu.be/-NN_m2yKAAk ) brought clarity to the problem.
Basically what I understood from the two is that time dilation and length contraction are not similar or symmetrical things (as I was assuming in original question) but by nature, we are able to measure time only using one and length using another method respectively.
In fact, there are concepts of length dilation and duration contraction that are not generally discussed but should be.
